It is given that y = (1 + 2cos x)cos 2x for 0 < x < π.
The complete set of values of x for which y is negative is
- A0 < x < π4, 2π3 < x < 3π4
- B0 < x < π4, 3π4 < x < π
- C0 < x < 2π3, 3π4 < x < π
- Dπ4 < x < 2π3, 3π4 < x < π
- Eπ4 < x < 2π3
- Fπ4 < x < 3π4
Show the answer and worked solution
answer · D
- A0 < x < π4, 2π3 < x < 3π4
- B0 < x < π4, 3π4 < x < π
- C0 < x < 2π3, 3π4 < x < π
- Dπ4 < x < 2π3, 3π4 < x < π
- Eπ4 < x < 2π3
- Fπ4 < x < 3π4
It is a product, so track the sign of each factor separately. 1 + 2cos x vanishes at cos x = −12, that is x = 2π3: it is positive to the left of that and negative to the right. For cos 2x, the angle 2x runs over (0, 2π), so cos 2x is positive on (0, π4) and (3π4, π) and negative on (π4, 3π4). The product is negative exactly where the signs differ: on π4 < x < 2π3 (positive times negative) and on 3π4 < x < π (negative times positive).